Why $\int \frac{dx}{\sqrt{9-x^2}} = sin^{-1}\frac{x}{3}$? I don't understand how the two formulas are equal since the right side involves trigonometric sine that the left is devoid of.
 A: You can use the following substitution
$$ (*) \qquad x=\dfrac{3}{2}\sin t \implies \text{ d}x = \dfrac{3}{2} \cos t \text{ d}t $$
This'll yield the following equality
$$ \underbrace{\int \dfrac{1}{\sqrt{9-4x^2}} \text{ d}x \ \overset{(*)}= \ \dfrac{1}{3} \int \dfrac{1}{|\cos t|} \dfrac{3}{2} \cos t \text{ d}t}_{\because \ \sqrt{9-4\left(\frac{9}{4}\sin (t)\right)^2} \ = \ 3 |\cos (t)|} \ = \underbrace{\dfrac{1}{2} \int \mathrm{sgn}\left(\cos(t)\right) \text{ d}t}_{\because \ \mathrm{sgn}\left(\cos(t)\right) \ \cdot \ |\cos(t)| \ =: \ \cos(t) }  = \dfrac{1}{2} \ t \ \mathrm{sgn}\left(\cos(t)\right) + \mathcal{C} $$
Finally, reverting our substitution back, we have that
$$ t = \arcsin \left( \dfrac{2x}{3} \right) $$
and that
$$ \mathrm{sgn}\left(\cos(t)\right) = \mathrm{sgn}\left( \sqrt{1-\left(\dfrac{2x}{3}\right)^2} \right) = +1 $$
so we finally have that
$$ \int \dfrac{1}{\sqrt{9-4x^2}} \text{ d}x \ = \ \dfrac{1}{2} \arcsin \left( \dfrac{2x}{3} \right) + \mathcal{C} $$
A: $Hint: $
What if you put $u = 3\sin t$ ?. Will the derivative and the expression kind of cancel out and make it easy ?
A: In general for every integral having $\sqrt{a^2-x^2}$ you can use the substitution $$u=a\sin x$$ and then differentiate and calculate the integral. $\sin^{-1} will& be appear.
A: My process of thinking when seeing an integral like this is the following:
$\int \frac{du}{\sqrt{1-u^2}}=\arcsin(u)+C$, where $C$ is an arbitrary constant, is a standard primitive. Going from this standard integral to the one given in the question isn't a big stretch really:
$\int \frac{dx}{\sqrt{9-x^2}}=\int \frac{dx}{3\sqrt{1-(\frac{x}{3})^2}}=\frac{1}{3}\int \frac{dx}{\sqrt{1-(\frac{x}{3})^2}}$.
Going from this we can use a variable substitution if we want to, or (as this isn't a very difficult integral) we could just think that the primitive must be looking like this:
$k\cdot \arcsin(\frac{x}{3})+C$, where $C$ again is arbitrary, and $k$ is a constant coefficient. To find the correct value for $k$ we can differentiate:
$\frac{d}{dx}(k\cdot \arcsin(\frac{x}{3})+C)=k\cdot\frac{1}{\sqrt{1-(\frac{x}{3})^2}}\cdot\frac{1}{3}=k\cdot\frac{1}{\sqrt{9-x^2}}$, so we see that if $k=1$ we get our integrand. From this we can reach the conclusion.
